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最新推荐文章于 2021-05-14 13:32:07 发布

追逐星辰的光

最新推荐文章于 2021-05-14 13:32:07 发布

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本文介绍了一个问题的解决方法：给定一系列长度，代表每个人可以向左攻击的距离，计算在同一轮攻击中（所有人都同时进行攻击）最终还能剩下多少人存活。

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Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length L_i. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - L_i.

You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 10⁶) — the number of guilty people.

Second line contains n space-separated integers L₁, L₂, ..., L_n (0 ≤ L_i ≤ 10⁹), where L_i is the length of the i-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Example

Input

4
0 1 0 10

Output

Input

2
0 0

Output

Input

10
1 1 3 0 0 0 2 1 0 3

Output

Note

In first sample the last person kills everyone in front of him.

题意：n个人，然后会输入n个人的刀长，刀有多长就可以杀死自己左边几个人，假如第n个人的刀长是3，那么n-1和n-2和n-3都将会被他杀死，问按照这种操作，同时杀人(此时注意，如果一个人被杀死，那么他也会杀死自己改杀死的人)，最后还活着几个人。

#include <stdio.h>
#include <string.h>
#define Max(a,b) ((a)>(b)?(a):(b))
int a[1000005];
int main() {
	int n, sum, s;
	while(~scanf("%d",&n)) {
		sum = 1;
		for(int i = 0; i < n; i++) {
			scanf("%d",&a[i]);
		}
		s = a[n-1];
		for(int i = n-2; i >= 0; i--) {
			if(s == 0) sum++;
			s = Max(s-1, a[i]);
		}
		printf("%d\n",sum);
	}
	return 0;
}